King's Gold

Many, many moons ago there was a fair king who oversaw 10 villages. As the king, he required a village tax of 1 gold bar per day per village. Each gold bar is supposed to weigh 1000g and have the village name etched in to it. However, the king has learned through an informant that one village has been cheating him the whole time and only giving him a bar that weighs 999g.

The king is furious and orders a scale to be brought in to test the weight of the gold bars. Assuming the king has access to all the gold bars ever sent to him, how can he figure out which village has been cheating him the whole time by using the scale only once?

Permalink: http://problemotd.com/problem/kings-gold/

Comments:

number the villages 0-9. Take ) gold bard from village 0, 1 from village 1, 2 from village 2 etc through to 9 from village 9.. and weigh the combined bars. You will be able to tell where the cheat village is by the final weight. If the final figure of the sum is 0, then village 0 was the cheat.
If 9 - Village 1 Cheated
If 8 - Village 2 Cheated
If 7 - Village 3 Cheated
If 6 - Village 4 Cheated
If 5 - Village 5 Cheated
If 4 - Village 6 Cheated
If 3 - Village 7 Cheated
If 2 - Village 8 Cheated
If 1 - Village 9 Cheated

Whoa!! I could have solved this but I made the assumption, since it was in ye olde times, that the scale did not give you an exact amount and that you had to compare one stack of gold to another for equivalence. I call shenanigans!! :)

We take 1 bar (999g or 1000g) from village 1.
We take 2 bars (2000g or 1998g) from village 2.

So our total of 1 + 2 bars can be either:
2998 grams or 2999 grams.

If we only have 1 bar that weighs 999, it will be 2999 (and village 1 cheated), but if we have 2998 then village 2 cheated.

This works on a larger scale as well when we take:
1 bar from village 1
2 bars from village 2
3 bars from village 3
4 bars from village 4
...
10 bars from village 10

Always try to break down the problem, because the number of villages is arbitrary. If we can figure out what one cheated out of two villages with only one weighing, than we can do it for any number of villages.

My solution seems much simpler than the rest of these, so that makes me assume I am completely wrong.

Take one bar from each of the 10 villages (We're assuming 9 aren't cheating at all and 1 is always cheating). Put 5 bars on one half of the scale, and 5 on the other. The side that has 4999g has the cheating village on it. Simply remove two bars at a time, one from each side (Noting which side had the correct weight). When the scales become even, you know that you just removed the cheating village's bar from the previously unequal side.